Question

Figure below shows two paths that may be taken by a gas to go from a state $A$ to a state $C$.

In process $AB,400J$ of heat is added to the system and in process $BC,100J$ of heat is added to the system. The heat absorbed by the system in the process $AC$ will be

• A. $300J$
• B. $380J$
• C. $500J$
• D. $460J$

Answer 1

Answer: (D)

According to given figure as shown below

Process $AB$ is isochoric,
i.e. $\Delta V=0$, so no work is done.
Heat added to the system in the process $AB$ is
$\Delta Q=400J$
According to first law of thermodynamics,
$\Delta Q=\Delta U+\Delta W$
where, $\Delta U$ is the change in internal energy and $\Delta W$ is the work done.
Since, $\Delta W=0$
$\Rightarrow \Delta U=\Delta Q=400J$
$\therefore$ Change in internal energy is $400J$.
Process $BC$ is isobaric and the work done is given by
$\Delta W=p\left(V_2-V_1\right)$
$=6\times 10^4\left(4\times 10^{-3}-2\times 10^{-3}\right)$
$=6\times 10^4\times 2\times 10^{-3}=120J$
Heat added to the system in the process $BC$ is
$\Delta Q=100J$
Since, $\Delta Q=\Delta U+\Delta W$
$\Delta U=\Delta Q-\Delta W=(100-120)J=-20J$
$\therefore$ Change in internal energy is $-20J$.
Total increase in internal energy is going from state $A$ to state $C$ is $400-20=380J$.
Work done in process $AC$ is the area under the curve. Area of the trapezium $ACDEA$
$=\frac{1}{2}\left(p_2+p_1\right)\times \left(V_2-V_1\right)$
$=\frac{1}{2}\left(6\times 10^4+2\times 10^4\right)\times \left(4\times 10^{-3}-2\times 10^{-3}\right)$
$=\frac{1}{2}\times 8\times 10^4\times 2\times 10^{-3}=80J$
Again, by using $\Delta Q=\Delta U+\Delta W$,
where $\Delta U=380J$ and $\Delta W=80J$
$\therefore \Delta Q=\Delta U+\Delta W$
$=380+80=460J$
Thus, the heat absorbed by the system in the process $AC$ will be $460J$.